Simplify and expand the following expression: $ \dfrac{1}{q + 9}- \dfrac{3}{q + 3}+ \dfrac{3}{q^2 + 12q + 27} $
Solution: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor the quadratic in the third term: $ \dfrac{3}{q^2 + 12q + 27} = \dfrac{3}{(q + 9)(q + 3)}$ Now we have: $ \dfrac{1}{q + 9}- \dfrac{3}{q + 3}+ \dfrac{3}{(q + 9)(q + 3)} $ The least common multiple of the denominators is: $ (q + 9)(q + 3)$ In order to get the first term over $(q + 9)(q + 3)$ , multiply by $\dfrac{q + 3}{q + 3}$ $ \dfrac{1}{q + 9} \times \dfrac{q + 3}{q + 3} = \dfrac{q + 3}{(q + 9)(q + 3)} $ In order to get the second term over $(q + 9)(q + 3)$ , multiply by $\dfrac{q + 9}{q + 9}$ $ \dfrac{3}{q + 3} \times \dfrac{q + 9}{q + 9} = \dfrac{3(q + 9)}{(q + 9)(q + 3)} $ Now we have: $ \dfrac{q + 3}{(q + 9)(q + 3)} - \dfrac{3(q + 9)}{(q + 9)(q + 3)} + \dfrac{3}{(q + 9)(q + 3)} $ $ = \dfrac{ q + 3 - 3(q + 9) + 3} {(q + 9)(q + 3)} $ Expand: $ = \dfrac{q + 3 - 3q - 27 + 3}{q^2 + 12q + 27} $ $ = \dfrac{-2q - 21}{q^2 + 12q + 27}$